Fast way to multiply a number by 7

Problem

Given an integer x, compute 7 * x efficiently using bitwise operations and additions/shifts.

Examples

Example 1

Input: x = 5
Output: 35

Example 2

Input: x = -3
Output: -21

Solution

Multiplying by small constants is often faster when expressed as shifts and adds. We show two compact methods: shift-and-subtract (recommended) and shift-and-add.

Method 1 — Shift and subtract (recommended)

Intuition

Observe that 7 = 8 - 1, and 8 * x is x << 3. So 7*x = (x << 3) - x. This uses a single left shift and one subtraction.

Approach

• Compute t = x << 3 (equivalent to x * 8).
• Return t - x.
• Use types that avoid overflow for large inputs if needed.

Code

C++

class Solution {
 public:
  long multiply_by_7(long x) {
    long t = x << 3; // x * 8
    return t - x;    // 8x - x = 7x
  }
};

Go

package main

func MultiplyBy7(x int64) int64 {
    t := x << 3 // x * 8
    return t - x
}

Java

class Solution {
    public long multiplyBy7(long x) {
        long t = x << 3; // x * 8
        return t - x;    // 7 * x
    }
}

Python

class Solution:
    def multiply_by_7(self, x: int) -> int:
        t = x << 3  # x * 8
        return t - x

Complexity

• ⏰ Time complexity: O(1) — constant number of bit/arith operations.
• 🧺 Space complexity: O(1).

Method 2 — Shift and add (alternative)

Intuition

Use 7 = 4 + 2 + 1, so 7*x = (x<<2) + (x<<1) + x — three shifts and two additions.

Approach

• Compute a = x << 2 (4x), b = x << 1 (2x), then return a + b + x.

Code

C++

class Solution {
 public:
  long multiply_by_7_alt(long x) {
    return (x << 2) + (x << 1) + x; // 4x + 2x + x
  }
};

Go

func MultiplyBy7Alt(x int64) int64 {
    return (x << 2) + (x << 1) + x
}

Java

class Solution {
    public long multiplyBy7Alt(long x) {
        return (x << 2) + (x << 1) + x;
    }
}

Python

class Solution:
    def multiply_by_7_alt(self, x: int) -> int:
        return (x << 2) + (x << 1) + x

Complexity

• ⏰ Time complexity: O(1).
• 🧺 Space complexity: O(1).